Problem: Is ${497353}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {497353}= &&{4}\cdot100000+ \\&&{9}\cdot10000+ \\&&{7}\cdot1000+ \\&&{3}\cdot100+ \\&&{5}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {497353}= &&{4}(99999+1)+ \\&&{9}(9999+1)+ \\&&{7}(999+1)+ \\&&{3}(99+1)+ \\&&{5}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {497353}= &&\gray{4\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {4}+{9}+{7}+{3}+{5}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${497353}$ is divisible by $3$ if ${ 4}+{9}+{7}+{3}+{5}+{3}$ is divisible by $3$ Add the digits of ${497353}$ $ {4}+{9}+{7}+{3}+{5}+{3} = {31} $ If ${31}$ is divisible by $3$ , then ${497353}$ must also be divisible by $3$ ${31}$ is not divisible by $3$, therefore ${497353}$ must not be divisible by $3$.